If there are 80 shirts - 30%are blue and 40%are red how many<br> are white

Question 1 2 pts Logic Puzzle: A certain clock runs fast; it indicates that 1 hour has passed when actually only 56 minutes have

Liono4ka [1.6K]

Using proportions, it is found that the correct time when it reads 10:30pm is of 9:40 pm.

<h3>What is a proportion?</h3>

A proportion is a fraction of total amount, and the measures are related using a rule of three.

In this problem, when 56 minutes have passed, the clock indicates that 60 minutes have passed. Hence, what is the time passed when the clock indicates that 12 hours and 30 minutes, that is, 60 x 12 + 30 = 750 minutes have passed?

Hence, the <em>rule of three</em> is:

56 minutes - 60 minutes

x minutes - 750 minutes

Applying cross multiplication:

$60x = 56 \times 750$

$x = \frac{56 \times 750}{60}$

$x = 700$

700 minutes is 660/60 = 11 hours + 40 minutes after 10:00 am, hence the correct time is of 9:40 pm.

You can learn more about proportions at brainly.com/question/24372153

7 0

3 years ago

The length of a rectangular pool is 20 feet and the width is 15 feet. The

Ray Of Light [21]

Answer:

$255

Step-by-step explanation:

As, the area of the pool bottom = length x breath

= 20 x 15 = 300 square foot

So for painting

Per square foot = $0.85

So for 300 square foot = $0.85 x 300

= $255

5 0

3 years ago

a. Fill in the midpoint of each class in the column provided. b. Enter the midpoints in L1 and the frequencies in L2, and use 1-

Tresset [83]

Answer:

$\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1} & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7} & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}$

Using the frequency distribution, I found the mean height to be 70.2903 with a standard deviation of 3.5795

Step-by-step explanation:

Given

See attachment for class

Solving (a): Fill the midpoint of each class.

Midpoint (M) is calculated as:

$M = \frac{1}{2}(Lower + Upper)$

Where

$Lower \to$ Lower class interval

$Upper \to$ Upper class interval

So, we have:

Class 63-65:

$M = \frac{1}{2}(63 + 65) = 64$

Class 66 - 68:

$M = \frac{1}{2}(66 + 68) = 67$

When the computation is completed, the frequency distribution will be:

$\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1} & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7} & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}$

Solving (b): Mean and standard deviation using 1-VarStats

Using 1-VarStats, the solution is:

$\bar x = 70.2903$